PDE for Heat Diffusion Equation

1. Mar 8, 2012

ZedCar

1. The problem statement, all variables and given/known data
The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

2. Relevant equations

3. The attempt at a solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.

A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

2. Mar 8, 2012

HallsofIvy

Staff Emeritus
That first "t" should be "T". I realize this is a typo.

Why?

Yes, but the second derivative with respect to t is irrelevant. Your differential equation has only the first derivative with respect to t.

Because that is the correct second derivative with respect to x, not t.

The first derivative with respect to t is a exp(at()cos(bx). If the differential equation ∂T/∂t= a∂^2T/∂x^2 were satisfied by exp(at)cos(bx) you would have to have
aexp(at)cos(bx)= -ab^2exp(at)cos(bt) which would imply a= -ab^2 which in turn gives 1= -b^2 so b^2= -1 which is impossible for real valued b.

3. Mar 8, 2012

ZedCar

Cheers for that! I see what I was doing wrong - getting x and t confused...

4. Mar 8, 2012

ZedCar

Another possible solution we investigated for the same heat diffusion equation was;

T(x,t) = exp(-at + ibx)

The PDEs being;

∂T/∂t = -a exp(-at + ibx)

∂^2 T / ∂x^2 = - b^2 exp(-at + ibx)

So the possible solution is possible if a = b^2 α

I understand all the above now.

Though apparently T(x, t) is complex in this instance. How is this apparent?

5. Mar 8, 2012

Ray Vickson

Be very glad that a, b > 0 does not give a solution, since such a "solution" would violate some basic laws of Physics: warm regions (where cos(bx) > 0) would get hotter, and cool regions (where cos(bx) < 0) would get colder, so heat would flow from cold to hot without any external influence---violating Thermodynamics. Also, warm regions would eventually get hotter than the hottest star, and cold regions would plunge well below absolute zero.

Look instead, at the case a < 0, b > 0. Does that behave properly?

RGV